Bromine water reacts with $S O_{2}$ to form

H_{2} O & H B r

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H_{2} S O_{4} & H B r

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H B r & S

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S & H_{2} O

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Solution

The correct option is D $H_{2} S O_{4} & H B r$
Bromine water reacts with $S O_{2}$ to form $H_{2} S O_{4} & H B r$

$B r_{2} + 2 H_{2} O + S O_{2} \to H_{2} S O_{4} + 2 H B r$

Suggest Corrections

Similar questions

B r_{2} + S O_{2} + H_{2} O \to H_{2} S O_{4} + H B r

H B r

H_{2} S O_{4}

H_{2} O + B r_{2} \to H O B r + H B r

H B r

The propene reacts with HBr to form

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